prove by induction that 4^n + 3^n +2 is a multiple of 3 ∀ n Z^+


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Question Number 73537 by Rio Michael last updated on 13/Nov/19

$p r o v e b y i n d u c t i o n t h a t 4^{n} + 3^{n} + 2 i s a m u l t i p l e o f 3$ $\forall n Z^{+}$
	Answered by mind is power last updated on 13/Nov/19
	$n=1 true 4+3+2=9 suppose ∀n∈N−{0} u_n =4^n +3^n +2 is multiple of 3 show 3∣u_(n+1) u_(n+1) =4.4^n +3^n ∗3+2 =3.4^n +4^n +3^n +2+2.3^n =4^n +3^n +2+3.4^n +3^n .2 =u_n +3(4^n +2.3^(n−1) ) since n≥1⇒3^(n−1) ∈N⇒3∣3.(4^n +2.3^(n−1) ) by hypothese 3∣U_n ⇒ 3∣u_(n+1)$
	$n = 1 t r u e$ $4 + 3 + 2 = 9$ $s u p p o s e \forall n \in N - {0} u_{n} = 4^{n} + 3^{n} + 2 i s m u l t i p l e o f 3$ $s h o w 3 ∣ u_{n + 1}$ $u_{n + 1} = 4 . 4^{n} + 3^{n} * 3 + 2$ $= 3 . 4^{n} + 4^{n} + 3^{n} + 2 + 2 . 3^{n}$ $= 4^{n} + 3^{n} + 2 + 3 . 4^{n} + 3^{n} . 2$ $= u_{n} + 3 (4^{n} + 2 . 3^{n - 1})$ $s i n c e n ⩾ 1 \Rightarrow 3^{n - 1} \in N \Rightarrow 3 ∣ 3 . (4^{n} + 2 . 3^{n - 1}) b y h y p o t h e s e 3 ∣ U_{n} \Rightarrow$ $3 ∣ u_{n + 1}$
	Commented by Rio Michael last updated on 13/Nov/19

	$t h a n k y o u s i r$
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